JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
If $\int_{0}^{1}\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=a+b\sqrt{2}+c\sqrt{3}$, where $a,b,c$ are rational numbers, then $2a+3b-4c$ is equal to :
Definite Integration
Solution
Concept. Rationalising turns the integrand into a difference of two square roots. Formulas used. (1) $\dfrac1{\sqrt{3+x}+\sqrt{1+x}}=\dfrac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)}=\dfrac{\sqrt{3+x}-\sqrt{1+x}}2$(2) $\int\sqrt{x+k}\,dx=\frac23(x+k)^{3/2}$Step 1 - Integrate. By (1), (2), $$\int_0^1=\frac13\Big[(3+x)^{3/2}-(1+x)^{3/2}\Big]_0^1=\frac13\Big[(8-2\sqrt2)-(3\sqrt3-1)\Big].$$ Step 2 - Simplify. $=3-\frac23\sqrt2-\sqrt3$, so $a=3$, $b=-\frac23$, $c=-1$. Step 3 - Value. $2a+3b-4c=6-2+4=8$. Answer: (C)